\newproblem{lay:6_3_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.3.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	You may assume that $\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3,\mathbf{u}_4\}$ is an orthogonal basis of $\mathbb{R}^4$. Let
	$\mathbf{u}_1=(0,1,-4,-1)$, $\mathbf{u}_2=(3,5,1,1)$, $\mathbf{u}_3=(1,0,1,-4)$, $\mathbf{u}_4=(5,-3,-1,1)$. Let $\mathbf{x}=(10,-8,2,0)$.
	Write $\mathbf{x}$ as the sum of two vectors, one in $\mathrm{Span}\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}$ and the other in
	$\mathrm{Span}\{\mathbf{u}_4\}$.
}{
   % Solution
	We project $\mathbf{x}$ onto $\mathrm{Span}\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}$
	\begin{center}
		$\begin{array}{rcl}
			\mathbf{x}_{123}&=&\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+
			                   \frac{\mathbf{x}\cdot\mathbf{u}_2}{\mathbf{u}_2\cdot\mathbf{u}_2}\mathbf{u}_2+
												 \frac{\mathbf{x}\cdot\mathbf{u}_3}{\mathbf{u}_3\cdot\mathbf{u}_3}\mathbf{u}_3\\
											&=&\frac{-16}{18}\begin{pmatrix}0\\1\\-4\\-1\end{pmatrix}+
											   \frac{-8}{36}\begin{pmatrix}3\\5\\1\\1\end{pmatrix}+
												 \frac{12}{18}\begin{pmatrix}1\\0\\1\\-4\end{pmatrix}
											 =\begin{pmatrix}0\\-2\\4\\-2\end{pmatrix}\\
			\mathbf{x}_4&=&\frac{\mathbf{x}\cdot\mathbf{u}_4}{\mathbf{u}_4\cdot\mathbf{u}_4}\mathbf{u}_4=\frac{72}{36}\begin{pmatrix}5\\-3\\-1\\1\end{pmatrix}=
			    \begin{pmatrix}10\\-6\\-2\\2\end{pmatrix}
		\end{array}$
	\end{center}
	It can be easily verified that $\mathbf{x}=\mathbf{x}_{123}+\mathbf{x}_4$.
}
\useproblem{lay:6_3_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
